Problem: Simplify the following expression: $y = \dfrac{-9x^2+16x- 7}{-9x + 7}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(-7)} &=& 63 \\ {a} + {b} &=& &=& {16} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $63$ and add them together. The factors that add up to ${16}$ will be your ${a}$ and ${b}$ When ${a}$ is ${7}$ and ${b}$ is ${9}$ $ \begin{eqnarray} {ab} &=& ({7})({9}) &=& 63 \\ {a} + {b} &=& {7} + {9} &=& 16 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 +{7}x) + ({9}x {-7}) $ Factor out the common factors: $ x(-9x + 7) - 1(-9x + 7)$ Now factor out $(-9x + 7)$ $ (-9x + 7)(x - 1)$ The original expression can therefore be written: $ \dfrac{(-9x + 7)(x - 1)}{-9x + 7}$ We are dividing by $-9x + 7$ , so $-9x + 7 \neq 0$ Therefore, $x \neq \frac{7}{9}$ This leaves us with $x - 1; x \neq \frac{7}{9}$.